Physics - Elasticity Question with Solution | TestHub

The length of a rod at temperature $T$ is $L=L_0(1+\alpha \Delta T)$, where $L_0$ is the initial length and $\Delta T$ is the change in temperature.

For the two rods, their lengths at temperature $T$ are:

$L_1={\ell }_1(1+{\alpha }_1\Delta T)$

$L_2={\ell }_2(1+{\alpha }_2\Delta T)$

The difference in lengths is $L_1-L_2={\ell }_1(1+{\alpha }_1\Delta T)-{\ell }_2(1+{\alpha }_2\Delta T)$.

$L_1-L_2={\ell }_1+{\ell }_1{\alpha }_1\Delta T-{\ell }_2-{\ell }_2{\alpha }_2\Delta T$

$L_1-L_2=({\ell }_1-{\ell }_2)+({\ell }_1{\alpha }_1-{\ell }_2{\alpha }_2)\Delta T$

For the difference in lengths to be independent of temperature, the term multiplied by $\Delta T$ must be zero.

So, ${\ell }_1{\alpha }_1-{\ell }_2{\alpha }_2=0$

${\ell }_1{\alpha }_1={\ell }_2{\alpha }_2$

Rearranging this gives:

$$\frac{{\ell }_1}{{\ell }_2}=\frac{{\alpha }_2}{{\alpha }_1}$$

The final answer is $\boxed{B}$.